/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 119 The blade divides the jet of wat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The blade divides the jet of water having a diameter of 3 in. If one-fourth of the water flows downward while the other three-fourths flows upward, and the total flow is \(Q=0.5 \mathrm{ft}^{3} / \mathrm{s}\), determine the horizontal and vertical components of force exerted on the blade by the jet, \(\gamma_{w}=62.4 \mathrm{lb} / \mathrm{ft}^{3}\)

Short Answer

Expert verified
The horizontal and vertical components of the force exerted on the blade by the jet are respectively \( 635.28 \, \text{lb} \) and \( 15.6 \, \text{lb} \).

Step by step solution

01

Calculate the cross-sectional area of the jet

First, determine the cross-sectional area of the jet. It's a circle so the formula will be \( A = \frac{\pi D^2}{4} \) where D is the diameter of the jet. Convert 3 inches to feet: \( 3 \, \text{in} = 0.25 \, \text{ft} \). Thus, \( A = \frac{\pi (0.25)^2}{4} = 0.0491 \, \text{ft}^2 \).
02

Calculate the weights of the upward and downward streams

The total flow rate is \( Q = 0.5 \, \text{ft}^3/\text{s} \). It's given that one-fourth flows downwards and three-fourths flows upwards. Therefore, the flow rate of the downward stream (\( Q_{down} \)) is \( 0.5/4 = 0.125 \, \text{ft}^3/\text{s} \) and the flow rate of the upward stream (\( Q_{up} \)) is \( 0.5*3/4 = 0.375 \, \text{ft}^3/\text{s} \). The weights of the downward and upward streams can be found by multiplying the flow rates by the unit weight of the water \( \gamma_{w} = 62.4 \, \text{lb/ft}^3 \). So, weight_down = \( Q_{down} * \gamma_{w} = 0.125*62.4 = 7.8 \, \text{lb} \) and weight_up = \( Q_{up} * \gamma_{w} = 0.375*62.4 = 23.4 \, \text{lb} \).
03

Calculate the horizontal and vertical components of the force on the blade

For horizontal component, since the water is deflected horizontally by 180°, the change in momentum is double. So, Momentum = Weight * Velocity. Using \( Velocity = Q/A = 0.5/0.0491 = 10.18 \, \text{ft/s} \), the horizontal force exerted on the blade is \( F_{x} = 2*(weight_up+weight_down)*Velocity = 2*(7.8+23.4)*10.18 = 635.28 \, \text{lb} \). For the vertical component, we only need to consider the weight of the water flowing upward. The downward force is \( -weight_down = -7.8 \, \text{lb} \). So, the net vertical force is \( F_{y} = 23.4 - 7.8 = 15.6 \, \text{lb} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Flow Rate Calculation
Understanding the concept of flow rate calculation is essential in fluid mechanics because it directly impacts how we analyze and predict fluid dynamics. In this exercise, flow rate is indicated as \( Q = 0.5 \, \text{ft}^3/\text{s} \). Flow rate represents the volume of fluid passing through a cross-section per unit time. It is a crucial measurement for describing the amount of fluid in motion.

Converting units, as required in the exercise, is often necessary. For example, converting inches to feet helps simplify calculations since other measurements like flow and density are typically given in feet and pounds. The relationship between diameter, area, and flow rate is important: the area (\( A \)) is derived from the diameter using the formula \( A = \frac{\pi D^2}{4} \).

To compute the upward and downward components of flow, knowing that \( \frac{1}{4} \) of flow goes downward and \( \frac{3}{4} \) goes upward allows us to distribute the total flow accordingly: \( Q_{down} = 0.125 \, \text{ft}^3/\text{s} \) and \( Q_{up} = 0.375 \, \text{ft}^3/\text{s} \). Being precise in these calculations helps define subsequent parameters in water jet dynamics.
Force Components
The analysis of force components requires understanding how fluid forces act in different directions. In this exercise, both horizontal and vertical forces are analyzed. These forces result from the interaction between the water and the blade.

**Horizontal Force**: Water hitting the blade and deflecting causes a change in momentum. If a fluid deflects horizontally, consider the direction change. Here, water is deflected at 180°, doubling the momentum change, calculated using the equation: momentum \( = \text{Weight} \times \text{Velocity} \). The overall horizontal force is then \( F_{x} = 2 \times (weight_{up} + weight_{down}) \times \text{Velocity} = 635.28 \, \text{lb} \). This force calculation considers the cumulative influence of both streams.

**Vertical Force**: Primarily based on the gravitational weight effects, it takes into account the net effect between upward and downward forces. The downward force negates part of the upward force, thus simplifying our calculation. The net vertical force becomes \( F_{y} = weight_{up} - weight_{down} = 15.6 \, \text{lb} \). This emphasizes the role of gravitational weight in vertical force analysis.
Water Jet Dynamics
Water jet dynamics explore how quickly and forcefully a jet of water interacts with surfaces. This exercise demonstrates the application of water dynamics on a blade.

**Velocity Calculation**: The velocity of the water jet propels how much force the water can exert on the surface. We've calculated it as \( \text{Velocity} = Q/A = 10.18 \, \text{ft/s} \). This is foundational in computing how vigorous the jet forces the blade, contributing to both vertical and horizontal forces.

**Jet Behavior and Momentum**: Water jets operate using principles of fluid dynamics, where the behavior of a flowing fluid involves momentum change. In 180° redirection, as seen with the horizontal deflection in this exercise, the complete reversal of flow direction implies a full momentum shift. It's vital to account for these directional changes in calculating resultant forces.

Understanding these dynamics helps in predicting how water jets interact with materials, which is invaluable for designing effective systems that utilize water pressure, such as turbines and pumps.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the \(10-\mathrm{kg}\) block which slides along the smooth surface and is subjected to a horizontal force of \(6 \mathrm{~N}\). If observer \(A\) is in a fixed frame \(x\), determine the final speed of the block in 4 s if it has an initial speed of \(5 \mathrm{~m} / \mathrm{s}\) measured from the fixed frame. Compare the result with that obtained by an observer \(B\), attached to the \(x^{\prime}\) axis that moves at a constant velocity of \(2 \mathrm{~m} / \mathrm{s}\) relative to \(A\).

A 0.03-lb bullet traveling at \(1300 \mathrm{ft} / \mathrm{s}\) strikes the 10-lb wooden block and exits the other side at \(50 \mathrm{ft} / \mathrm{s}\) as shown. Determine the speed of the block just after the bullet exits the block. Also, determine the average normal force on the block if the bullet passes through it in \(1 \mathrm{~ms}\), and the time the block slides before it stops. The coefficient of kinetic friction between the block and the surface is \(\mu_{k}=0.5\).

The \(20-\mathrm{g}\) bullet is traveling at \(400 \mathrm{~m} / \mathrm{s}\) when it becomes embedded in the \(2-\mathrm{kg}\) stationary block. Determine the distance the block will slide before it stops. The coefficient of kinetic friction between the block and the plane is \(\mu_{k}=0.2\)

A 1-Ib ball \(A\) is traveling horizontally at \(20 \mathrm{ft} / \mathrm{s}\) when it strikes a \(10-\) lb block \(B\) that is at rest. If the coefficient of restitution between \(A\) and \(B\) is \(e=0.6\), and the coefficient of kinetic friction between the plane and the block is \(\mu_{k}=0.4\), determine the time for the block \(B\) to stop sliding.

The nozzle discharges water at a constant rate of \(2 \mathrm{ft}^{3} / \mathrm{s}\). The cross-sectional area of the nozzle at \(A\) is \(4 \mathrm{in}^{2}\) and at \(B\) the cross-sectional area is \(12 \mathrm{in}^{2}\). If the static gauge pressure due to the water at \(B\) is \(2 \mathrm{lb} / \mathrm{in}^{2}\), determine the magnitude of force which must be applied by the coupling at \(B\) to hold the nozzle in place. Neglect the weight of the nozzle and the water within it. \(\gamma_{w}=62.4 \mathrm{lb} / \mathrm{ft}^{3}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.